a) PTHH: \(Mg+2HCl\rightarrow MgCl_2+H_2\)
b) Ta có: \(n_{Mg}=\frac{2,4}{24}=0,1\left(mol\right)\)
\(\Rightarrow n_{H_2}=0,1mol\) \(\Rightarrow V_{H_2}=0,1\cdot22,4=2,24\left(l\right)\)
c) Theo PTHH: \(n_{HCl}=2n_{Mg}=0,2mol\)
\(\Rightarrow m_{HCl}=0,2\cdot36,5=7,3\left(g\right)\) \(\Rightarrow m_{ddHCl}=\frac{7,3}{20\%}=36,5\left(g\right)\)
d) Ta có: \(n_{MgCl_2}=n_{Mg}=0,1mol\) \(\Rightarrow m_{MgCl_2}=0,1\cdot97=9,7\left(g\right)\)
Mặt khác: \(m_{H_2}=0,1\cdot2=0,2\left(mol\right)\)
\(\Rightarrow m_{dd}=m_{Mg}+m_{ddHCl}-m_{H_2}=2,4+36,5-0,2=38,7\left(g\right)\)
\(\Rightarrow C\%_{MgCl_2}=\frac{9,7}{38,7}\cdot100\approx26,06\%\)